Understanding the Concept:
In a redox balancing calculation, the total number of equivalents or electrons lost during oxidation must equal the total number of equivalents or electrons gained during reduction:
\[
\text{Moles of Reductant} \times v\text{-factor}_{\text{reductant}} = \text{Moles of Oxidant} \times v\text{-factor}_{\text{oxidant}}
\]
The valence factor ($v$-factor) represents the net change in oxidation state per mole of the substance.
Step 1: Determine the $v$-factor for the reduction of dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$).
In an acidic medium, the dichromate ion is reduced to chromium(III) ions:
\[
\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^2 + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
\]
The oxidation state of $\text{Cr}$ changes from $+6$ to $+3$. Since each dichromate formula unit contains 2 chromium atoms, the net electron gain is:
\[
v\text{-factor for }\text{K}_2\text{Cr}_2\text{O}_7 = 2 \times (6 - 3) = 6
\]
Step 2: Determine the $v$-factor for the oxidation of ferrous oxalate ($\text{FeC}_2\text{O}_4$).
Ferrous oxalate contains two oxidizable parts: the ferrous cation ($\text{Fe}^{2+}$) and the oxalate anion ($\text{C}_2\text{O}_4^{2-}$). Both are oxidized under acidic conditions:
$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (Electron loss $= 1$)
$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$ (Electron loss $= 2$)
The total number of electrons lost per mole of $\text{FeC}_2\text{O}_4$ is:
\[
v\text{-factor for }\text{FeC}_2\text{O}_4 = 1 + 2 = 3
\]
Step 3: Equate electron transfer values to find $x$.
Given $1\text{ mole}$ of ferrous oxalate reacts with $x\text{ moles}$ of $\text{K}_2\text{Cr}_2\text{O}_7$:
\[
1 \times 3 = x \times 6 \implies x = \frac{3}{6} = 0.5\text{ moles}
\]